You could set this up as a caldculaus problem and find the maximum-income number of buses, or just calculate and compare the income for each the three choices given.
With 10 buses, you have 10x32 = 320 m^2 for buses and 805-320 = 485 m^2 for 485/5 = 97 cars. Total income is 60 + 194 = $254 . However, the number of vehicles exceeds tha allowed .
With 15 buses, you have 480 m^2 for buses and 325 m^2 for 65 cars. The income is 90+130 = $220
Now consider the case of 28 buses. Chances are the total income wil be less than the 15 bus case.
The area of a parking lot is 805 square meters. A car requires 5 square meters and a bus requires 32 square meters of space. At most 80 vehicles can park at one time. If the cost to park a car is $2.00 and the cost to park a bus is $6.00, how many buses should be in the lot to maximize income?
I'm given the choices 10, 15, or 28
2 answers
15 buses